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Proof step 19 from proof abs-paths-eq (previous step)

tree[p3 + p2].dir?, tree[p3 + p].dir?, abs(tree, ino0)(dirs), ispath(p3 + p, ino0, ino1, dirs, files), ispath(p3 + p2, ino0, ino1, dirs, files), p ∈ tree[p3], p2 ∈ tree[p3], p3 ∈ tree, ino1 ∈ dirs, p2 ≠ ε ∧ p ≠ ε → p2.head ≠ p.head, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ p = ε, p2 = p, p2 = ε

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
tree[p3 + p].dir?, tree[p3 + p2].dir?, abs(tree, ino0)(dirs), ispath(p3 + p2, ino0, ino1, dirs, files), ispath(p3 + p, ino0, ino1, dirs, files), ino1 ∈ dirs, p3 ∈ tree, p2 ∈ tree[p3], p ∈ tree[p3], ino0 ∈ dirs, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ p2.head = p.head, p = ε, p2 = ε

Rule argument

Additional information

Used simplifier rules: path-in-dirs: p ≠ ε ⊦ ispath(p, n, m, dirs, files) → n ∈ dirs; (forward) from specification afs-base%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ p + p0 = ε ↔ p = ε ∧ p0 = ε; (as)